dokku ls shows the following but how can I get the status of app-1470418443 using awk for example ?
So that magic command returns running.
-----> App Name Container Type Container Id Status
app-1470418443 web 78a092d176f1 running
example web 3b7803c49e04 running
example2 web a750f2b4be44 running
awk '$1=="app-1470418443"{print $NF}'
Since there is trailing space in the example output you have to get all non-space characters ([^ ]*) and then all the space (_*, underscore _ represents space as space didn't render at all) in the end of the string ($) and print only the matching parts (grep -o):
$ grep "^app-1470418443 " file | grep -o "[^ ]* *$"
running
Without trailing space, the latter grep would simplify a bit (well 2 bytes) to match only the non-space chars:
$ grep "^app-1470418443 " file | grep -o "[^ ]*$"
running
Related
I am trying to use awk to print the unique lines returned by a command. For simplicity, assume the command is ls -alh.
If I run the following command in my Z shell, awk shows all lines printed by ls -alh
ls -alh | awk '!seen[$0]++'
However, if I run the same command with $SHELL -c while escaping the ! with backslash, I only see the first line of the output printed.
$SHELL -c "ls -alh | awk '\!seen[$0]++'"
How can I ensure the latter command prints the exact same outputs as the former?
EDIT 1:
I initially thought the ! could be the issue. But changing the expression '!seen[$0]++' to 'seen[$0]++==0' has the same problem.
EDIT 2:
It looks like I should have escaped $ too. Since I do not know the reason behind it, I will not post an answer.
In the second form, $0 is being treated as a shell variable in the double-quoted string. The substitution creates an interestingly mangled awk command:
> print $SHELL -c "ls -alh | awk '\!seen[$0]++'"
/bin/zsh -c ls -alh | awk '!seen[-zsh]++'
The variable is not substituted in the first form since it is inside single quotes.
This answer discusses how single- and double-quoted strings are treated in bash and zsh:
Difference between single and double quotes in Bash
Escaping the $ so that $0 is passed to awk should work, but note that quoting in commands that are parsed multiple times can get really tricky.
> print $SHELL -c "ls -alh | awk '\!seen[\$0]++'"
/bin/zsh -c ls -alh | awk '!seen[$0]++'
I'm writing a shell script which shut down some services and trying to get its pid by using the following awk script.
However, this awk script can't get pid. What's wrong with that?
ps -ef | awk -v port_no=10080 '/[m]ilk.*port=port_no/{print $2}'
The result of ps -ef is like this:
username 13155 27705 0 16:06 pts/2 00:00:00 /home/username/.rbenv/versions/2.3.6/bin/ruby /home/username/.rbenv/versions/2.3.6/bin/milk web --no-browser --host=example.com --port=10080
This process is working with a different port argument as well, so I want to kill the process only working on port=10080.
The awk script below works fine, but when I specify the port no using awk -v like the above, it doesn't work well.
ps -ef | awk '/[m]ilk.*port=10080/{print $2}'
awk version: GNU Awk 4.0.2
The syntax for pattern matching with /../ does not work with variables in the regular expression. You need to use the ~ syntax for it.
awk -v port_no=10080 '$0 ~ "[m]ilk.*port="port_no{print $2}'
If you notice the regex carefully, the regex string on the r.h.s of ~ is under the double-quotes ".." except the variable name holding the port number which shouldn't be under quotes, for the expansion to happen.
This task is easily accomplished using pgrep:
$ pgrep -f '[m]ilk.*port=10080'
Have a look at man pgrep for details.
I am trying to read some files located on a server and write only a certain number of columns from those files onto my local machine. I am tried to do this in a for loop to avoid inputting my password for each file. Below is what I was able to cobble till now.
The following code works but writes all the output to a single file which is not manageable due to its large size.
ssh user#xx.xxx.xxx.xx 'for loc in /hel/insur/*/201701*; do zcat $loc | grep -v NUMBER | awk -F',' -v OFS="," '\''{print $1,$2,$3,$4,$5}'\'' | gzip; done' > /cygdrive/c/Users/user1/Desktop/test/singlefile.csv.gz
So, I tried to write each file individually as shown below but it gives me an error saying that it cannot find the location(possibly because I am sshed ito the remote server).
ssh user#xx.xxx.xxx.xx 'for loc in /hel/insur/*/201701*; do zcat $loc | grep -v NUMBER | awk -F',' -v OFS="," '\''{print $1,$2,$3,$4,$5}'\'' | gzip > /cygdrive/c/Users/user1/Desktop/test/`echo $loc | cut -c84-112` ; done'
Any ideas on how to solve this?
I need to find the ID of some container docker, but some containers have similar names:
$ docker images
REPOSITORY TAG IMAGE ID
app-node latest 620350b79c5a
app-node-temp latest 461c5143a985
If I run:
$ docker images | grep -w app-node-temp | awk -e '{print $3}'
461c5143a985
If I run instead:
$ docker images | grep -w app-node | awk -e '{print $3}'
620350b79c5a
461c5143a985
How can I match the exact name?
I'd say just use awk with exact string matching:
docker images | awk '$1 == "app-node" { print $3 }'
Dashes are considered non-word characters, so grep -w won't work when the difference is marked by a dash.
In context, grep '^app-node[[:space:]]' would work. It looks for the required name followed by a space.
Of course, grep | awk is an anti-pattern most of the time; it would be better to use:
docker images | awk '/^app-node[[:space:]]/ { print $3 }'
Or, an easier solution with awk again uses equality — as suggested by Tom Fenech in his answer:
for server in app-node app-node-temp
do
docker images | awk -v server="$server" '$1 == server { print $3 }'
…
done
If running docker images is too expensive, you can run it once and capture the output in a file and then scan the file. This shows how to pass a shell variable into the awk script.
The chances are the pipeline would be run to capture the container's image ID information:
image_id=$(docker images | awk -v server="$server" '$1 == server { print $3 }')
docker images -q is good for your case
i ran the following command from console it output the correct result:0,
sudo -H -u hadoop bash -c "/home/hadoop/hadoop-install/bin/hadoop dfsadmin -report | grep 'Under replicated blocks' | awk '{print \$4}'"
however if i put it in shell script and assigned it to a variable the 'awk' won't work anymore, it just output the whole result from 'grep':
replications=`sudo -H -u hadoop bash -c "/home/hadoop/hadoop-install/bin/hadoop dfsadmin -report | grep 'Under replicated blocks' | awk '{print \$4}'"`
echo "Replications: $replications"
result: Replications: Under replicated blocks: 0
how can i make the awk work again to only output 4th column which is 0 instead of the whole string?
In backtick command substitution, \ followed by $ means just $. From the POSIX standard:
Within the backquoted style of command substitution, backslash shall retain its literal meaning, except when followed by: '$', '`', or '\' (dollar sign, backquote, backslash). (...)
With the $(command) form, all characters following the open parenthesis to the matching closing parenthesis constitute the command. Any valid shell script can be used for command, except a script consisting solely of redirections which produces unspecified results.
And yet more explicitly from the bash manpage:
When the old-style backquote form of substitution is used, backslash retains its literal meaning except when followed by $, `, or \. The first backquote not preceded by a backslash terminates the command substitution. When using the $(command) form, all characters between the parentheses make up the command; none are treated specially.
So the easiest way is to use
replications=$(sudo -H -u hadoop bash -c "/home/hadoop/hadoop-install/bin/hadoop dfsadmin -report | grep 'Under replicated blocks' | awk '{print \$4}'")
but
replications=`sudo -H -u hadoop bash -c "/home/hadoop/hadoop-install/bin/hadoop dfsadmin -report | grep 'Under replicated blocks' | awk '{print \\\$4}'"`
also works.